Definition. A path on a topological space X is a continuous map The path is said to connect x and y in X if f(0)=x and f(1)=y. X is said to be path-connected if any two points can be connected by a path.

In a sense, path-connectedness is more active since one requires an explicit path to establish it, while the earlier connectedness is more passive since it simply indicates a failure to decompose as a topological disjoint union. The two are obviously related, starting from:

Theorem 1. A path-connected space X is connected.

Proof.

Suppose X is path-connected but not connected. There’s a surjective continuous map Pick such that f(x)=0 and f(y)=1. There’s a path such that g(0)=x and g(1)=y. Now the composition is continuous and surjective, which contradicts the fact that [0, 1] is connected. ♦

The converse is not true: the topologist’s sine curve is connected but not path-connected. Let where

Let be projection maps to the x– and y-coordinates respectively. Then contains 0 and 1, so its image is the whole [0, 1] by the intermediate value theorem. Hence, Pick points such that

Since [0, 1] is compact, f is uniformly continuous. So there exists ε>0 such that whenever satisfy we have Since [0, 1] is compact, we’ll pick m<n such that By intermediate value theorem, there’s a point u between tm and tn such that Then but which is a contradiction.

[ Notice it took quite a bit of effort to prove a seemingly obvious claim, and we needed compactness to prove it. ]

Note also that X is closed in R2, and every point in Y is a point of accumulation of Z, so cl(Z) = X. In short, we have the first bummer.

Conclusion. The closure of a path-connected subset Y of X is not necessarily path-connected.

Thankfully, the next result does carry over.

Proposition 2. If is a continuous map of topological spaces and X is path-connected, then so is f(X).

Proof.

For any we can pick such that Pick a path such that and Then the composition gives a path which connects to ♦

Proposition 3. If is a collection of path-connected subspaces of X and then so is

Proof (Sketchy).

Pick If then and for some indices i and j. Since Yi and Yj are path-connected and contain x, there’s a path in Yi connecting x to y and a path in Yj connecting x to z. Hence, concatenating gives a path connecting y to z. ♦

Proposition 4. If is a collection of path-connected spaces, then is also path-connected.

Proof.

Let Since each is path connected, pick a path such that Now let be the path

To check that f is continuous, let’s use the universal property of products. It suffices to show is continuous for each i, where is the projection map. But so we’re done. ♦

Note

Once again, this fails for the box topology on Since the example in the previous article is not connected, it cannot be path-connected.

Path-Connected Components

As before, we obtain the concept of path-connected components. We define, for points x, y in X, a relation x ~ y if and only if they belong to some path-connected component. Proposition 3 then tells us this gives an equivalence relation.

Definition. The equivalence classes of the above-mentioned relation are called the path-connected components.

Since a path-connected component is automatically connected, each connected component is a disjoint union of path-connected components.

Examples

It’s easy to see that any interval (closed, open or half-open) is path-connected. In particular, it’s connected.

Hence is a disjoint union of and each of which is a path-connected component.

Consider Q as a subspace of R. Since the connected components are singleton sets, the path-connected components can’t break them down any further.

Take the topologist’s sine curve above. Y and Z are both path-connected since they’re homeomorphic to intervals. Since X is not path-connected, the path-connected components must be Y and Z. Note that Y is open while Z is closed in X. This is one example where connected components decompose further into path-connected components.